Mechanics of Materials Solutions Chapter07 Probs13 23

7/25/2019 Mechanics of Materials Solutions Chapter07 Probs13 23

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7.13 For the cantilever beam and loading shown,(a) Derive equations for the shear force V and the

bending moment M for any location in the beam.

(Place the origin at point A.)(b) Plot the shear-force and bending-moment

diagrams for the beam using the derived functions.

Fig. P7.13

Solution

Beam equilibrium:

Section a-a: For the interval 0 ≤ x < 8 ft:

Section b-b: For the interval 8 ft ≤ x < 14 ft:

( )( )

( )( )( )

5 kips/ft 6 ft 0

30 kips

120 kip-ft 5 kips/ft 6 ft 3 ft 0

210 kip-ft

y y

y

C C

C

F C

C

M M

M

Σ = − =

∴ = −

Σ = + + =

∴ = −

-

0 0 kips

120 kip-ft 0 120 kip-ft

y

a a

F V V

M M M

Σ = − = ∴ =

Σ = + = ∴ = −

( )( )

( )( )-

2

5 kips/ft 8 ft 0

5 40 kips

8 ft120 kip-ft 5 kips/ft 8 ft =0

2

2.5 40 280 kip-ft

y

b b

F x V

V x

x M x M

M x x

Σ = − − − =

∴ = − +

−⎛ ⎞Σ = + − +⎜ ⎟

⎝ ⎠

∴ = − + −





(b) Shear-force and bending-moment diagrams:





7.14 For the cantilever beam and loading shown,(a) Derive equations for the shear force V and the

bending moment M for any location in the beam.

(Place the origin at point A.)(b) Plot the shear-force and bending-moment

diagrams for the beam using the derived functions.

Fig. P7.14

Solution

Beam equilibrium:



( )( )

( )( )( )

( )( )

16 kips/ft 9 ft 17 kips 0

2

44 kips

16 kips/ft 9 ft 6 ft

2

17 kips 14 ft 0

400 kip-ft

y y

y

A A

A

F A

A

M M

M

⎛ ⎞Σ = − − =⎜ ⎟

⎝ ⎠

∴ =

⎛ ⎞Σ = − − ⎜ ⎟

⎝ ⎠

− =

∴ = −

( )

2

1 6 1 644 kips 0

2 9 2 9

44 kips3

1 6

2 9 3

1 6 400 kip-ft 44 kips 0

2 9 3

y y

a a A y

x x F A x V x V

xV

x x M M A x x M

x x x x M

−

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞Σ = − − = − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

∴ = − +

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞Σ = − − + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= − + + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦3

44 400 kip-ft9

x M x∴ = − + −

( )( )

( )( )

1 6 kips/ft 9 ft2

1 44 kips 6 kips/ft 9 ft 0

2

17 kips

y y F A V

V

V

⎛ ⎞Σ = − −⎜ ⎟⎝ ⎠

⎛ ⎞= − − =⎜ ⎟

⎝ ⎠

∴ =





( )( )( )

( ) ( )( )( )

-

16 kips/ft 9 ft 6 ft

2

1 400 kip-ft 6 kips/ft 9 ft 6 ft 044 kips

2

17 238 kip-ft

b b A y M M A x x M

x x M

M x

⎛ ⎞Σ = − − + − +⎜ ⎟

⎝ ⎠

⎛ ⎞= − + − + =⎜ ⎟

⎝ ⎠

∴ = −






7.15 For the simply supported beam subjected to theloading shown,

(a) Derive equations for the shear force V and the

bending moment M for any location in the beam.(Place the origin at point A.)

(b) Plot the shear-force and bending-moment

diagrams for the beam using the derived functions.(c) Report the maximum positive bending moment,

the maximum negative bending moment, and their respective locations.

Fig. P7.15

Solution

Beam equilibrium:



( )( )( )

( )

( )( )( )( )

250 kip-ft 7 kips/ft 25 ft 12.5 ft

17 ft 0

113.97 kips

7 kips/ft 25 ft

113.97 kips 7 kips/ft 25 ft 0

61.03 kips

A

y

y

y y y

y

y

M

C

C

F A C

A

A

Σ = −

+ =

∴ =

Σ = + −

= + − =

∴ =

( ) ( )

( )

( ) ( )

2

61.03 kips 07 kips/ft 7 kips/ft

7 61.03 kips

7 kips/ft2

061.03 kips 7 kips/ft2

3.5 61.03 kip-ft

y y

a a y

F A x V x V

V x

x M A x x M

x x x M

M x x

−

Σ = − − = − − =

∴ = − +

⎛ ⎞Σ = − + +⎜ ⎟⎝ ⎠

⎛ ⎞= + + =− ⎜ ⎟

⎝ ⎠

∴ = − +

( ) ( )

( )

( ) ( )

2

61.03 kips 07 kips/ft 7 kips/ft

7 61.03 kips

250 kip-ft7 kips/ft2

250 kip-ft 061.03 kips 7 kips/ft2

3.5 61.03

y y

b b y

F A x V x V

V x

x M A x x M

x x x M

M x

−

Σ = − − = − − =

∴ = − +

⎛ ⎞Σ = − + + +⎜ ⎟

⎝ ⎠

⎛ ⎞= + + + =− ⎜ ⎟

⎝ ⎠

∴ = − + 250 kip-ft x −





Section c-c: For the interval 17 ft ≤ x < 25 ft:


(c) Maximum bending moment and its

location

M max-positive = 266.04 kip-ft

@ x = 8.72 ft

M max-nagative = –224.0 kip-ft

@ x = 17 ft

( )

( )

( )

( ) ( ) ( )

7 kips/ft

61.03 kips 113.97 kips 07 kips/ft

7 175 kips

( 17 ft) 250 kip-ft7 kips/ft2

(17 ft)175 kips 113.97 kips 7 kips/ft2

y y y

c c y y

F A C x V

x V

V x

x M A x C x x M

x x x

−

Σ = + − −

= + − − =

∴ = − +

⎛ ⎞Σ = − − − + + +⎜ ⎟⎝ ⎠

= − + +

2

250 kip-ft 0

3.5 175 2,187.5 kip-ft

M

M x x

⎛ ⎞+ + =⎜ ⎟

⎝ ⎠

∴ = − + −





7.16 Use the graphical method to construct theshear-force and bending-moment diagrams for the

beams shown. Label all significant points on each

diagram and identify the maximum moments (both positive and negative) along with their respective

locations. Clearly differentiate straight-line and

curved portions of the diagrams.Fig. P7.16

Solution

Beam equilibrium:

Shear-force and bending-moment diagrams:

( )( ) ( )( ) ( )34 kips 4 ft 56 kips 8 ft 14 ft 0

41.71 kips

34 kips 56 kips

41.71 kips 34 kips 56 kips 0

48.29 kips

A y

y

y y y

y

y

M D

D

F A D

A

A

Σ = − − + =

∴ =

Σ = + − −

= + − − =

∴ =










Solution

Beam equilibrium:


( )( ) ( )( )

( )( ) ( )

20 kN 4 m 30 kN 8 m

10 kN 14 m 10 m 0

46 kN

20 kN 30 kN 10 kN

46 kN 20 kN 30 kN 10 kN 0

14 kN

A

y

y

y y y

y

y

M

D

D

F A D

A

A

Σ = − −

− + =

∴ =

Σ = + − − −

= + − − − =

∴ =









curved portions of the diagrams.

Fig. P7.18

Solution

Beam equilibrium:


( )( ) ( )( )

8 kips 18 kips 0

10 kips

8 kips 8 ft 18 kips 2 ft 0

28 kip-ft

y y

y

C C

C

F C

C

M M

M

Σ = − + + =

∴ = −

Σ = − + =

∴ = −










Fig. P7.19

Solution

Beam equilibrium:


( )( )( ) ( )

( )( )

( )( )

12 ft 6 ft 18 ft 010 kips/ft

40 kips

10 kips/ft 12 ft

40 kips 10 kips/ft 12 ft 0

80 kips

A y

y

y y y

y

y

M C

C

F A C

A

A

Σ = − + =

∴ =

Σ = + −

= + − =

∴ =










Fig. P7.20

Solution

Beam equilibrium:


( )( )( ) ( )

( )( )

( )( )

4.5 kips/ft 12 ft 6 ft 9 ft 0

36 kips

4.5 kips/ft 12 ft

36 kips 4.5 kips/ft 12 ft 0

18 kips

A y

y

y y y

y

y

M B

B

F A B

A

A

Σ = − + =

∴ =

Σ = + −

= + − =

∴ =










Fig. P7.21

Solution

Beam equilibrium:


( )( )

( )( )( )

4 kN/m 3 m 0

12 kN

4 kN/m 3 m 1.5 m 10 kN-m 0

28 kN-m

y y

y

A A

A

F A

A

M M

M

Σ = − =

∴ =

Σ = − − − =

∴ = −





7.22 Use the graphical method to construct the

shear-force and bending-moment diagrams for the





Solution

Beam equilibrium:


( )( )

( )( ) ( )( ) ( )

28 kips 9 kips/ft 5 ft 0

17 kips

28 kips 8 ft 9 kips/ft 5 ft 2.5 ft 0

111.5 kip-ft

y y

y

C C

C

F C

C

M M

M

Σ = − + =

∴ =

Σ = − + + =

∴ =





7.23 Use the graphical method to construct the

shear-force and bending-moment diagrams for the





Solution

Beam equilibrium:


( )( )( ) ( )

( )( )

10 kips/ft 9 ft 4.5 ft 6 ft 20 ft 0

47.25 kips

10 kips/ft 9 ft 0

42.75 kips

A y

y

y y y

y

M D

D

F A D

A

Σ = + + =

∴ = −

Σ = + + =

∴ = −

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Mechanics of Materials Solutions Chapter07 Probs13 23